# Is it a Wishart matrix?

We know that an $m \times m$ random matrix $\boldsymbol{A} = \boldsymbol{H} \boldsymbol{H}^H$ is a (central) real/complex Wishart matrix with $n$ degrees of freedom and covariance matrix $\boldsymbol{\Sigma}$ $(\boldsymbol{A} \sim \mathcal{W}_m (n, \boldsymbol{\Sigma}))$, if the columns of the $m \times n$ matrix $\boldsymbol{H}$ are zero mean independent real/complex Gaussian vectors with covariance matrix $\boldsymbol{\Sigma}$.

If the columns of matrix $\boldsymbol{H}$ are zero mean independent real/complex Gaussian vectors with different covariance matrix, i.e, the $i$-th column of $\boldsymbol{H}$ has covariance matrix $\boldsymbol{\Sigma}_i$ ($i=1, \cdots, n$), is matrix $\boldsymbol{A}$ a Wishart matrix, and what is its distribution? Thanks!

It is not a Wishart random matrix. When $m=1$ then ${\cal W}_1(n,\sigma^2) = \sigma^2 \chi^2_n= \Gamma\left(\frac{n}{2},\frac{1}{2\sigma^2}\right)$ is the distribution of $\sum_{i=1}^n {X_i}^2$ where $X_i\sim_{\text{iid}} {\cal N}(0,\sigma^2)$.
Taking $X_i\sim{\cal N}(0,\sigma_ i^2)$ with different $\sigma_i$'s yields a sum of independent Gamma distributions with different scale parameters, and this is not a Gamma distribution in general.
• Thank you for your help! Does your answer mean that when $\boldsymbol{A}$ is not a Wishart matrix when $m = 1$, thus it is not a Wishart matrix when $m > 1$? – lala Jul 10 '14 at 8:12
• Because $\boldsymbol{A}$ is a really matrix in my problem ($m > 1$), so does it have any way to prove in general case? Thanks – lala Jul 10 '14 at 8:18
• @lala My answer shows this is not true for $m=1$. For $m>1$, one can show that the same fact occurs for the diagonal entries of the matrix: the diagonal entries of a random Wishart matrix are Gamma random variables, but they are not Gamma anymore if you use different $\Sigma$'s. – Stéphane Laurent Jul 10 '14 at 9:07
• @ Stéphen Laurent: Could I ask you one more question? How can I prove the diagonal entries of matrix $\boldsymbol{A}$ are not Gamma random variables if using the different $\boldsymbol{\Sigma}$? Must I use the pdf? (It is difficult way, isn't it?) – lala Jul 10 '14 at 11:02
• @lala Consider the $(1,1)$-entry of the random matrix. If you take different values of the $(1,1)$-entry of $\Sigma$, then this is exactly the situation when $m=1$. – Stéphane Laurent Jul 10 '14 at 11:21